Question 162945
How high did the frog jump?
{{{y = -0.25x^2+5x}}}
This is a quadratic equation and, when graphed, you will see a parabola that opens downward. The dependent variable (y) represents the height of the frog's jump at time (x) seconds.
You may see this kind of equation written as:
{{{h(t) = -0.25t^2+5t}}}
The vertex of the parabola is, of course, a maximum, and this represents the maximum height of the frog's jump.
So you need to find the coordinates of the vertex.
The x-coordinate of the vertex is given by:{{{x = -b/2a}}} where the a and b come from the general form of the equation:
{{{ax^2+bx+c}}} In your problem, a = -0.25 and b = 5, so...
{{{x = -5/2(-0.25)}}}
{{{x = 10}}} now substitute this value of x into the original quadratic equation to find the corresponding y-coordinate.
{{{y = -0.25(10)^2+5(10)}}} Simplify and solve for y.
{{{y = -0.25(100)+50}}}
{{{y = 25}}}
The frog's maximum height is 25 feet.
{{{graph(400,400,-2,24,-5,30,-0.25x^2+5x)}}}